c h a t r motion in two and three dimensions · in describing motion in two and three dimensions,...

C H A T R

Motion in Two andThree Dimensions

6.1 INTRODUCTION

So long as we are dealing with one-dimensional motion, rectangular (Cartesian) coordinateshave proved quite satisfactory. In describing motion in two and three dimensions, however, rec-tangular coordinates are frequently not very helpful or convenient, especially in demonstratingphysical significance in a given situation. For that purpose we introduce other coordinate sys-tems, such as plane polar, cylindrical, and spherical polar. Furthermore, we will discuss differ-ent vector operators in these coordinate systems, and express kinematics and potential energyfunctions in them. To gain a physical-conceptual understanding of situations such as harmonicoscillators in two and three dimensions and projectile motion, we will need to use these coor-dinate systems.

6.2 DIFFERENT COORDINATE SYSTEMS

To describe the position and motion of an object or a point in space, it is necessary to have a co-ordinate system. Some of the commonly used coordinate systems are rectangular coordinates,plane polar coordinates, cylindrical coordinates, and spherical polar coordinates.

Rectangular or Cartesian Coordinates

To start, we choose a two-dimensional rectangular coordinate system. It consists of two mutu-ally perpendicular coordinate axes crossing at the origin O, as shown in Fig. 6.1. As shown, theX- and F-axes are in the plane of the paper and at 90° to each other. The position of point P is

188

Sec. 6.1 Different Coordinate Systems 189

+ Y

o^ Figure 6.1 Rectangular coordinates

+X (x, y) of a point P in two dimensions.

described by the coordinates (x, y), which are obtained by drawing perpendiculars (or projec-tions) from P to the X- and F-axes, so that OA = x and OB = y. Thus we may write

OP2 = OA2 + OB2 = x (6.1)

Figure 6.2 shows a set of three-dimensional rectangular coordinate axes. Again the X-and F-axes are in the same plane and at 90° to each other, while the Z-axis is perpendicular tothis plane. Once again the position of point P is described by the coordinates (x, y, z), and wemay write

OP2 = OM2 + OC2 = (OA2 + OB2) + OC2

OP2 = x2 + y2 + z2 (6.2)

The three mutually perpendicular axes shown in Fig. 6.2 form a right-handed system.

or

I''X, y, z)

Figure 6.2 Rectangular coordinates(x, y, z) of a point P in three dimen-

190

Plane Polar Coordinates

Motion in Two and Three Dimensions Chap. 6

Rectangular coordinate systems are quite useful in describing the motion of an object movingin a straight line. Such coordinates are not always useful when the motion is curved, as in cir-cular motion. For such motion other suitable coordinates are used. A proper choice of a coordi-nate system can make problem solving much simpler. For example, circular motion in a planeis best described by plane polar coordinates.

Referring to Fig. 6.3, the rectangular coordinates of point P in the XY plane are (x, y). PointP is located at a distance r from the origin, and the line OP makes an angle 8 with the X-axis. Itis equally acceptable to describe the position of point P by the coordinates (r, ff), called planepolar coordinates. The relations between (x, y) and (r, ff) from Fig. 6.3 are

x = r cos 8, y = r sin (6.3)

We can express r and 8 in terms of x and y by a simple procedure. By squaring and addingEqs. (6.3), we get

x + y = r2(cos 8 + sin2 8) = r2

Also from Eq. (6.3), we get

That is

r sin 6

r sin 8= tan 6

r = Vx2 + y2 and 0 = t a n - 1 - (6.4)

Thus, in a two-dimensional coordinate system, (x, y) or (r, 8) completely specify the position ofa point in a plane, r can have any value between 0 and <», while 8 can have any value between0 and 277 radians, with 8 increasing counterclockwise.

Further comparison and contrast between rectangular and plane polar coordinates aredemonstrated in Figs. 6.4 and 6.5. Figure 6.4 shows the plots of constant x and constant y, while

>. Figure 6.3 Plane polar coordinates+X (r, 0) of a point P in two dimensions.

x = constant

A" O

I i

y = constant

rCartesian coordinates:

for each line y = constant,while x variesfor each line x = constant,while y varies

Figure 6.4 Plots of constant x and constant y. For each continuous line, y = con-stant while x varies: for each dashed line^x = constant while y varies.

Fig. 6.5 shows the plots of constant r and constant 6. In Fig. 6.4, plots of x = constant are straightlines parallel to the F-axis, while plots of y = constant are straight lines parallel to the X-axis.These two sets of lines are perpendicular to each other as shown. In Fig. 6.5 plots of 8 = con-stant are straight lines starting from the origin O and directed radially outward, while plots ofr = constant are circles with their centers at the origin. Again, the lines of 8 = constant and r =constant are perpendicular to each other wherever they cross.

Cylindrical Coordinates

Let us consider a point P located at a distance r from the origin O. Point P can be located byusing a set of rectangular coordinates (x, y, z) or cylindrical coordinates (p, </>, z), as shown inFig. 6.6 and explained next, or by means of spherical polar coordinates (r, 8, <f>), as discussedlater.

Cylindrical coordinates (p, 4>, z) are shown in Fig. 6.6 and are related to (x, y, z) by theequations

x = p cos <f> (6.5a)

192 Motion in Two and Three Dimensions Chap. 6

8 = constant

Plane polar coordinatesfor each line 8 = constant,while r variesfor each line r = constant,

while 9 varies

Figure 6.5 Plots of constant r and constant 6. For each continuous line, 6 =constant while r varies; for each dashed line r = constant while 6 changes, (f and8 are unit vectors replacing i and j.)

+Z

P(x, y, z)

+ Y

Figure 6.6 Cylindrical coordinates(p, 4>, z) of a point P in space.

y = p sin

z = z

(6.5b)

(6.5c)

while the reverse relations can be obtained from Eqs. (6.5a) and (6.5b) by the procedure usedfor plane polar coordinates. That is,

- Vx2+ (6.6a)

* = t a n - (6.6b)

2 = z (6.6c)

Note that in cylindrical coordinates p has replaced r and <f> has replaced 6 of plane polarcoordinates.

Spherical Polar Coordinates

Once again consider point P in space located at a distance r from the origin O, as shown inFig. 6.7. The rectangular coordinates of the point P are (x, y, z), while its spherical polar coor-dinates are (r, 6, 0). To find the relation between these two sets of coordinates, we first resolveOP = r into two components PM and OM, where

PM= OC= OP cos 9 or z= r cos 9

OM = PC = OP sin 9 or OM = r sin 9

Z A

= r sin 0 cos <f>

y = r sin 0 sin 0 Figure 6.7 Spherical polar coordi-nates (r, 0, 0) of a point P in space.

We further resolve OM into two components, OA and OB, such that

OA = OM cos cf> or x = r sin 6 cos </>

OB = OM cos </> or y = r sin 0 sin </>

Thus we have the relations

x = r sin 6 cos </> (6.7a)

y = rsin 8 sin (/> (6.7b)

z = r cos 0 (6.7c)

The reverse relations, that is, r, 6, <fi in terms of x, y, z, may be obtained from Fig. 6.7 as follows:

r = OP = VOM2 + OC2 = V(OA2 + OB2) + OC2 = V*2 + y2 + z2

PC OM VOA2 + OB2 V r 2 + y2

tan0 =

tan <f> =

OC OC OCOB y

OA x

That is, we have the relations

r = Vx 2 + y2 + z2 (6.8a)

tan 0 = - ' ~ ' •/- (6.8b)z

tan ^ = - (6.8c)x

These spherical coordinates will be quite useful in discussing motion in three dimensions.

6.3 KINEMATICS IN DIFFERENT COORDINATE SYSTEMS

We are interested in describing the motion of a particle without regard to the forces that producesuch motion. Thus we shall describe the position, velocity, and acceleration of a particle in twoand three dimensions. The different coordinate systems that we use in describing motion in de-tail are plane polar coordinates, cylindrical coordinates, and spherical coordinates.

Cartesian Coordinates

The position of a particle P in the XY plane may be described by the coordinates (x, y), or pointP may be described by means of a position vector r = (x, y), where r is the distance from a spec-ified point called the origin. The motion of the point P in the XY plane may be given by de-scribing y as a function of x, or vice versa; that is,

y = y(x), x = x(y) (6.9)

Sec. 6.3 Kinematics in Different Coordinate Systems

or it will be still better to give a functional relation between x and y, such as

fix, y) = 0

For example, a particle moving in a circular path may be described by

x2 + y2 = a2

195

(6.10)

(6.11)

where a is the radius of the circle.A most convenient way to represent the path of a particle is in terms of some parameter s,

such as

x = x(s), y = y(s)

or r = r(s)

In such situations, a particle moving in a circular path will be described by

x = a cos 6, y = a sin 6

where 6 is the parameter in this case.

In general, if a particle moves in a plane XY, its motion is described by

x = x{t), y = y(t)

or r = r(f)

(6.12)

(6.13)

(6.14)

(6.15a)

(6.15b)

where time t is the parameter in this case. We may write the position vector r, in terms of unitvectors, as

r = IX + jy

The velocity and the acceleration of the particle and their components are

dr , dx , dy ~ *

dy d2r ^d2x ^d2y

A three-dimensional motion is represented as

+ j + kz

dr * dx ~ dy ~ dz ~— = i — + j — + k = iidt dt dt dtd\ d2r * Sx d2z

a =

(6.16)

(6.17)

(6.18)

(6.19)

(6.20)

(6.21)


Figure 6.8 Rectangular coordinates(x, y) and plane polar coordinates (r, 6)of a point P. Also shown are the unitvector i and j .

Plane Polar Coordinates

In many situations it is convenient to use plane polar coordinates (r, 6) instead of rectangularcoordinates (x, v) to describe the motion of a particle. The relations between the two sets of co-ordinates (Fig. 6.8) are

x = rcos

while the reverse relations are

r = (x1 + y1)"2

y = r sin

= tan '( —) = cos 1 = sin"1 -

(6.22)

(6.23a)

(6.23b)+ / / Wx2 + y2)

The distance r is measured from the origin, while the polar angle 6 is measured counterclock-wise from the X-axis, as shown in Fig. 6.8. Unit vector i and j in rectangular coordinates are asshown. We now define two unit vectors in polar coordinates that are perpendicular to each other.These two unit vectors are r and 6 and are in the direction of increasing r and 0, as shown inFig. 6.9. Thus r points in the direction of P along the increasing radial distance r, while 0 points

Figure 6.9 Unit vector f and 8 inplane polar coordinates.

Sec. 6.3 Kinematics in Different Coordinate Systems 197

j cos 6 II j sin 8I

O

- 4 -- i sin 9 icos 0

Figure 6.10 Relation between unitvectors (f, 9) and (i, j).

in the direction that P would move as angle 9 increases. Furthermore, both unit vectors are func-tions of angle 6. Unit vectors r and 8 form a new coordinate system called plane polar coordi-nates or simply polar coordinates. As shown in Figs. 6.9 and 6.10, unit vectors r and 8 are re-lated to \ andj by the relation

r = i cos 6 + j sin 9

8 = - 1 sin 9 + j cos 9

Let us differentiate these with respect to 9; that is,

dr ^ /j. /v— = — l sin 9 + j cos 9 = 6d9

(6.24)

(6.25)

Thus we have

— = — i cos 9 — j sin 9 = — rdd

dr a dQ— = 6 and — = — rd9 d9

(6.26)

These results can be obtained directly by referring to Fig. 6.1 l(a) and (b). These figures showthe positions of r and 8 for a particular angle 9 and 9 + d9. As angle 9 increases by d9, the ra-dial unit vector changes from r(0) to f (9 + d9) by an amount dr . Similarly, the angular unitvector changes from §(0) to §(0 + d9) by an amount c?9 as shown. Note that dr points in thedirection of 6, while J6 points in the direction opposite to r that is, in the direction of — r asshown.

The position vector r in terms of polar coordinates is given by

r = rr = rr(9) (6.27)


8(0 +dO)

o(a)

O

(b)

Figure 6.11 For calculating the variation of (a) f with 6 and (b) § with d.

Note that r = r(0); hence expression (6.27) does not contain 0 explicitly. The motion of theparticle is determined by r(t) and d(t) in polar coordinates. Thus the velocity v is

dr d dr dxv = — = — (rr) = — r + r —

dt dt dt dtSince r = f (ff) using Eq. (6.26) we must write

dr dr d6

dt dd dt= 90

That is,

We may identify

v = rr + ri

v=r and va = rd

(6.28)

(6.29)

where vr is the component of the velocity along r and is called the radial velocity, while ve isthe component along 6 and is called the angular velocity.

The acceleration of the system is given by

d\ d . , • , dr A . dr dd dr • A d6 R • dQ dda = — = —(rr + rOQ) = — r + r + — 08 + r — 6 + rd

dt dt dt dd d d ' '" '

= rr + r(6)0 + r06

+dd dt dt

r0(-r)0

dt dO dt

That is,

a = (r - r02)? + (rd + 2r0)0 (6.30)

Thus the two components of the acceleration a are the radial acceleration ar and the angular ac-celeration a0 given by

a = r - rd1

ae = rd + 2rd

(6.31)

(6.32)

Sec. 6.3 Kinematics in Different Coordinate Systems

A few remarks are in order at this time. The term

199

= 7 (6.33)is the centripetal acceleration arising from the motion in the 0 direction. Furthermore if ris held constant in time, r = r = 0, the path is a circle with centripetal accelerationar- r6 - - xf-Jr. The term 2r6 in ag is the Coriolis acceleration, and we shall postpone itsdiscussion to Chapter 11.

Cylindrical Polar Coordinates

By adding a Z component to plane polar coordinates we get cylindrical coordinates for de-scribing motion in three dimensions. The three unit vectors p, $, and z in the direction of in-creasing p, 0, and z, respectively,^ shown in Fig. 6.12. It is important to note that z is con-stant, while the unit vector p and <|> are functions of 0 as in the case of plane polar coordinates

The relations between rectangular coordinates (x, y, z) and cylindrical coordinates (p 

y = p sin (f>

z = z

(6.34a)

(6.34b)

(6.34c)

Figure 6.12 Cylindrical coordinates (p, <f>, z) and the corresponding unit •tors (p ,$ ,z) .

200

while the inverse relations are

P = (x2

) and

and, as before,


+ y2y2

1 y • - i y - i— — 91t1 — (~*(~\Q3111 / tUo

x Vx2 + y2

with an additional Z component,

p = i cos = — I sin </> + j cos (/>

J p _ .

dcp

X

Vx2 +

we may

y2

write

(6.35a)

(6.35b)

the relations

(6.36a)

(6.36b)

(6.37)

^ • = - p (6.38)dcp

The position vector r describing the location of a point P in cylindrical coordinates, shownin Fig. 6.12, is

r = pp + zz (6.39)

where p gives the distance of P from the Z-axis and <p gives its angular rotation from the X-axis,while z gives its elevation above the XY plane. Thus we may write the velocity vector, keepingin mind that p = p($) , as

dx _dy ~ dt~ dt

where dz/dt =

PP

0;

_l_ A \

zz

hence

_ dp~ dt

V = pp

dp dcp

' d&

dz^dtZ

+ zz

dzZ dt

(6.40)

Similarly, a = — = — (pp + pep 2)p + (p<£ + 2p<^)$ + zz (6.41)

Now we can express any vector A in terms of three components Ap, A^, and Az in the di-rections of the three mutually perpendicular unit vector p , <J>, and z. That is,

A = App+ A J> + Azz (6.42)

where the components depend not only on the vector itself but also on its location in space. Thisis because both p and . Thus if A is a function of a parameter, say t, then in eval-uating dA/dt we must remember the variation of p and <}> as demonstrated next. UsingEq. (6.42),

dA

dt

dA,, dp dq> d(f> dAz ^* d(f> dt + dt Z +

since dz/dt = O,dp/d(j) = q>, and dq>/d(f> = — p , we get, after rearranging,

dAj,

dtAz

dzz dt

dA

Spherical Polar Coordinates

dA(6.43)

Spherical polar coordinates or spherical coordinates are the most commonly used coordinatesin situations of spherical symmetry—for example, in the case of coulomb forces in atoms andgravitational forces. The point P in space is located by the coordinates (r, 6, </>), as shown inFig. 6.13. r is the radial distance from the origin O, 4> is the azimuthal angle locating a planewhose angle of rotation is measured from the X-axis, while angle 6 is the polar angle measureddown from the Z-axis. The polar angle 6 can have any value between 0 and TT/2, while the az-imuthal angle 4> can have any value between 0 and IT.

(a) (b)

Figure 6.13 (a) Spherical polar coordinates (r, 6, <£) and the corresponding unitvectors (f, 8, <J>) fl>) Orientation of unit vectors (f, §, <$>) relative to the coordi-nate system XYZ and polar angle 6.

Rectangular coordinates (x, y, z) are related to spherical polar coordinates (r, 6, <f>) by thefollowing relations (see Fig. 6.13):

(6.44a)

(6.44b)

(6.44c)

X

y

z

= r sin 9 cos

= r sin 8 sin

= r cos 9

Note that r sin 8 = p. The reverse relations are

r =

0 =

4> -

(x2 + y2 +

. V*2-!toritan

z

tan"' *

- v 2

(6.45a)

(6.45b)

(6.45c)

The three mutually perpendicular unit vectors used in spherical polar coordinates arer, 6, and $ , as shown in Figs. 6.13(a) and(b). Also shown are the unit vectors t , j , k, z (= k),and p . The unit vector <J> lies in the XY plane, while r , S, p , and z all lie in one vertical plane.For fixed r and 8 the variation in (f>, corresponds to rotation about the Z-axis, while for fixed rand 4> the variation in 8 corresponds to the rotation in the plane containing r , 9, p , and z. FromFig. 6.13 we can write the following relations between the unit vectors:

r = p sin 9 + z cos 6 = i sin G cos </> + j sin 6 sin 4> + k cos 6

8 = p cos 8 — z sin 9 = i cos 6 cos <f> + j cos 6 sin <f> — ft cos 8

= - 1 sin cos

(6.46a)

(6.46b)

(6.46c)

Differentiating these equations, we obtain the following relations:

dr , dr

88 d(f>= q> sin

38= $ cos (6.47)

— - = 0 — - = —p = — r sin 0 — 0 cos 8B8 dq>

These relations can also be derived from geometrical considerations by drawing figures similarto the ones in the case of plane polar coordinates.

In spherical coordinates, the position of a point P in space is given by the position vec-tor r:

r = rr = rr(8, </>) (6.48)

We can now find expressions for velocity and acceleration by making use of the preceding re-lations. Thus

dr d dr ^ dr drv = r = — = — [rr(0, <f>)] = -r + r — = rr + r—~

dt dt dt dt dt

Using

Hence

Eqs

we

• (6.47),

obtain

dr~~r (0> 4dt

M - ^ ,d*^ d<}>

d<f>lit

v = rr + r0§ + {r<j> sin 0)$ (6.49)

Similarly,

.. d\ d .„a = r = —- = — [rr + rb

dt dt- (r<f> s in

aj ai

which on simplification yields

a = (r - r02 - rsin2 0</>2)r + (rd + 2r0 - rsin 0cos 0</>2)S

+ (r sin 0 <j>' + 2'r<j) sin 0 + 2rd <j> cos ^ (6.50)

Since f, 6, and 0 form a set of mutually perpendicular unit vectors, we may write anyvector A in component form as

A=Arr (6.51)

where the components depend upon not only on vector A, but also on its location in space. If Ais a function of parameter t, that is, a function of time, we may write

JA dAr ^ dr dA^ A. dQ dA, «

dt dt r dt dt e dt dt

As before, using Eqs. (6.47) we may write

dr dr dd dr d<f> ~ • t

— = " ^ -7- + ~ 7 7 j = 80 + 0 sin 0 </)dt dd dt d<{> dt

dt

dt dd dt

dip dip d(b

dt dq> dt= - P<A = ( - ? sin 0 - 0 cos 9)<f>

Using these results in the preceding equation, we obtain

- Aee - A^ineA + {^f + Are - A.cose~

dt+ Ar sin 8 </> + Ae cos 6 <f> cj> (6.52)

6.4 DEL OPERATOR IN CYLINDRICAL AND SPHERICALCOORDINATES

We are now in a position to express the del operator in cylindrical and spherical coordinates byusing the definition of gradient. In cylindrical coordinates, a scalar function u is

Therefore, knowing

and

we may write

Using the relations

U = U(p, (j), Z)

du du dudu = — dp H d(j> H dz

dp dq> dz

r = pp + zz

dx = dpp + p — d(f> + dz zdq>

dpa n d

d<|>

we obtain

dr = p dp + <J>p dcf) + z dz

The definition of grad u is

du = Vu • dr = dr • Vu

For this relation to yield Eq. (6.54) we must define

„ ~ d - 1 d „ dV ^ p — + $ + z

dp p d</> dz

(6.53)

(6.54)

(6.55)

(6.56)

(6.57)

(6.58)

(6.59)

Sec. 6.5 Potential Energy Function

Thus, if we use Eqs. (6.59) and (6.57) in Eq. (6.58), we get Eq. (6.54); that is,

dr • Vu = (p dp + $pd(f> + z dz) • p — + <J>- — + z —\ dp p d dz

Similarly, for spherical coordinates, and using Eq. (6.48),

u = u(r, 6, 4>)

du , du dudu = — dr + — dd + — dd>

dr de d<f) *

205

ind if we define

dr = r dr + Qrdd + <j>rsin 6dcf>

„ d . 1 3 ~ 1V = r + 6 + <}>

dr r dd r sin 6

(6.60)

(6.61)

(6.62)

(6.63)

:t will satisfy du = dr • VM, as it should.

5 5 POTENTIAL ENERGY FUNCTION

In considering the motion of a particle in one dimension, we defined a potential energy functionis Eq. (2.77),

V(x) = f ' F(x) dx = - f F(x) dxx xs

and the corresponding force F(x) is

(6.64)

Fix) = -dV(x)

dx(6.65)

We can now extend these ideas to the motion of a particle in three dimensions.Let us consider a particle at r(x, y, z) that under the action of force F moves from I1! to r2.

The work done is given by

Wfr2

= F(r) dr (6.66)

and to evaluate this integral we must specify a path. As in the case of one dimension, we can in-troduce a potential energy function V(r) = V(x, y, z) as the work done by the force when theforce moves the particle from point r to some standard reference point rs. That is,

V(r) = f F(r) • dr = - f F(r) • dr (6.67)

Knowing that V(r) must be a function of position alone, this definition is possible only if thepreceding integral is independent of the path of integration, that is, only if F(r) is a conserva-tive force so that the work done will be independent of the path; hence the potential energy func-tion may be introduced.

Our task is now to proceed to find the necessary and sufficient conditions for F(r) to beconservative, and hence justification for the existence of a potential function V(r). Let us saythat work done in going from point P to Q is independent of the path. This means that the workdone in a closed path (see Fig. 6.14) in going from P to Q and back to P is zero. That is, we find

P->Q->P closedpath

F • dr = 0

According to Stokes' theorem [Eq. (5.141)], we may write this result as

W, closedpath

F - dr = / / « • < * x F) dS = 0surface

This can be true only if the integrand itself V X F is zero; that is

V x F = curl F = 0

(6.68)

(6.69)

(6.70)

This is a necessary and sufficient condition for the force to be conservative; hence a potentialenergy function given by Eq. (6.67) will exist. Thus Eq. (6.70) states a necessary and sufficientcondition for the existence of a potential function. A force F for which the curl F is zero is calleda conservative force.

Figure 6.14 The work done in a con-servative force field for a closed path ingoing from P to Q and back to P iszero.

Sec. 6.5 Potential Energy Function 207

We can now show that the existence of a potential function leads to the conservation oftotal energy if the force field is conservative. The work done by F in acting through Ti to r2 maybe written as

2 F(r) • dr = f * F(r) • dr + [= I F(r) • dr + | F(r) • dr = V,(r) - V2(r) (6.71)

But the work done is also equal to the change in kinetic energy

- K, (6.72)

Combining Eqs. (6.71) and (6.72),

V,(r) = K2 + V2{r)

That is, if E is the total energy, we obtain

K+V= \m{x2 + y2 + z2) + V(x, y,z) = E

•.vhich is the energy integral for motion in three dimensions.Let us consider F = F(r, t), and say at any time t that

V x F(r, 0 = 0

and we can define a potential function

dr

>o that F(r,0 = -

(6.73)

(6.74)

(6.75)

(6.76)

(6.77)

But in such cases the sum of the kinetic and potential energy is not constant; hence F(r, t) isnot a conservative force.

If F is given and we want to evaluate V(r), we can proceed either directly by evaluating:he line integral,

(6.78)

?T we can evaluate three ordinary

Fx - -

V(r) = - f F(r)

integrals by starting

T' Fy ~ ~dx ydV

dv'

• dr

with

77Fz -

dV

dz

208

to obtain


V= -

V= -j Fydy + C2(x,z)

V= ~^

(6.79)

The resulting potential function must be consistent with all three expressions given inEqs. (6.79). This is demonstrated in the following example.

y Example 6.1

Show that the following forces are conservative and find the corresponding potentials.

(a) F = ax\ + by} + cA(b) Fx = 3ayz3 - 20to3y2, Fy = 3axz2 - l0bx4y, F2 = 9axz2y

Solution

In order to prove that the force representsa conservative force field, we must provethat curl of the force vector is zero.

(a)dx dy

k:=l

dz

Sas

i

A

ax

j

B

b-y

k

C

c-z

( i)

Let i, j , and k be the unitvectors. A, B, and Crepresent the differentialoperators as shown,(a) Let Sa = Curl F,which is shown in matrixform. After calculating theabsolute value of Sa and S a = i — ( c z ) - i - - ( b y ) - — ( j c ) z - t - — ( k b ) y + — ( a x j ) (a-x-k)

d d d d d d

Sa=i-B-c-z- i-C-b-y- A-j-c-z-|-A-k-b-y-i-a-x-j-C- a-x-k-B

i idy

then substituting differentialoperators for A, B, and Sa_0

C, we simplify and findthatSa = 0. Thus Fin thiscase is a conservativeforce field.

Using Eq. (6.78) or (6.79), we cancalculate the potential Vacorresponding to this conservativeforce as shown.

dz

Vaa

(dx

(dx

(dz

(dy

(-a-x)dx-i- -b-ydy-h | -c-zdz

. , - l 2 1 . 2 l 2V a = — - a - x -b-y c-z

2 2 2

Va=—a-x -b-y c-z M-Constant2 2 2

(ii)


The constant is equal to the sum of three constants Ca = Cx + Cy + Cz.

The three constants Cx, Cy, and Cz, corresponding to the three components of the force, arecalculated from the initial conditions.

i b) We follow the same procedure as in (a) and prove that the force is conservative.

Sb=

1

A

J

B

k 1C

3-a-y-z - bx y 3-a-x-z - 10-b-x y 9-a-xz y(iii)

2 3 4 25b=9-i-B-a-x-z y - 3-i-C-a-x-z + 10-i-C-b-x - y - 9-A-j-a-x-z -y-t-i ...

+ 3-A-k-a-x-z3 - 10A-k-b-x4-y-H 3-a-yz3-j-C - 3-a-yz3-k-B - b-x3-y2-j-C + b-x3-y2-k-B

Sb=O

3 2- ( 3 - a - x - z - 1 0 - b - x -y) d y V z = - ( 9 - a - x - z -y) dz- ( 3 - a - y - z - 2 0 - b - x -y dx V y =

3 4 2 3 4 2 3. x=-3-a-y-z-x-t-5-b-x-y -|-Cx Vy=-3-a-y-z -x-i-5-b-x -y -t- Cy Vz=-3-a-y-z-x-i-Cz (iv)

For a conservative force field, if we assumeCx = Cy = 0, Vz is short of a term andCz must be as shown.

4 2Cx=0 Cy=0 Cz(x,y)=5-b-x -y

Hence for a conservative force, the potential is

Vb=-3-a-y-z -x-f-5-b-x -y -t-Cx-|- (-3-a-y-z -x-i-5-b-x -y + Cyj -+- (-3-a-y-z -x-i-Czj

4 2 3Vb=5-b-x -y - 3-a-x-y-z

Vb=- 9-a-yz3-x+ 10-b-x4-y2 + Cx +• Cy •+- Cz(v)

EXERCISE 6.1 Show that the following forces are conservative and calculate the correspondingrotentials.

a I F = — yzi — xzj — xyti.b) Fx = -6ayzx, F = az(z2 - 3x2), Fz = 3ay(z2 - x2)

) Example 6.2

\ particle of mass m moves from point A to B around a semicircular path of radius R, as shown inr:g. Ex. 6.2. It is attracted toward its starting point A by a force proportional to its distance from A. When


it reaches B, the force toward A is Fo. Calculate the work done against this force when the particle movesfrom A to B in this semicircular path as shown.

Solution

The work done is given by

rB

= F --M

W =

and F = kr

where k is a constant. But when r = 2R at B (see Fig. Ex. 6.2),

FB = Fo = k(2R) or k =

rp

Therefore, F = - £ r

To express r in terms of R and 0, we use the law of cosines:

r2 = R2 + R2 - 2R2 cos 6 = 2R2(l - cos 6)

2R

F = kr = r£«V2(l - cos 0)2

Figure Ex. 6.2

From the Fig. Ex. 6.2, we get the relationsbetween different quantities given here.

ds«R-de

cos(\(/)=cos(180- (|))=cos((|))=cos —

The work W done in moving from A to B is W= Fds

(i)

(iii)

(iv)


Note that

Fds|=F-r-cos(y)-d9=F-r-cos — and . F 0

*2RR-j2-(l-cos(9))

Substituting for F, ds, and cos(v|/), the workWF done can be calculated as shown. W F = FO

R-J2-(l-cos(9))-R-cos2-R

d9

WF=-FOR

EXERCISE 6.2 Solve the problem by the energy conservation method by noting the fact that, for aoring, F = kr and potential energy is jkr .

/ Example 6.3

A particle moving in an XY plane as shown in Fig. Ex. 6.3 is attracted toward the origin by a force F =.. v. Calculate the work done when the particle moves (a) from A to B and then to C, and (b) from A to C^ong an elliptical path given by the equations x = 2a sin 6 and y = a cos 6.

Solution

a) In going from A to B and to C, the force is F = kly, and the work done is given by

B cF • ds + F • ds= f F • rfs = [ F - d s + [ F •

-M -M ->B

B C

= F dr cos 0, + F dr cos 02•M •'«

ror the path from A(0, a) to 5(2a, a),

y = a, dr = dx, cos 0, = —

r ?r the path from B(2a, a) to C(2a, 0)

x = 2a, dr = cfy, cos 02 =

zor both the paths

V(2a)2+

(i)

(H)

(iii)

(iv)


The minus sign can also be justified by the fact that the force is opposing the displacement.

Yk Yk

A(0, a) B(2a, a) a

C(2a, 0) x la X

(a) (b)

Figure Ex. 6.3

r2-a

Substituting Eqs. (ii), (iii), and (iv) in (i),the work Wa done in going from A to Band then to C is as shown.

Wa= - k - - dx-i-

+a

-k

(2-a)2+y2

-dy

Wa=k- U 5 - 1 - lnU5+l

(b) We now calculate the work done along the elliptical path. The equation of the ellipse is

(2a)2 a2

and r2 = x2 + y2

In terms of the inscribed and circumscribed circles and the central angle 0

x = 2a sin 0 and y = a cos 0

The work done is going from A to C along this elliptical path is

W= f F • dr = [ F, dx + j" Fy dy

where, from Eq. (vii),

dx = 2a cos Odd, dy = -a sin 0d0

x k x kFx = F cos (f) = F - = = —

2a sin 9

r y r a cos 6 V(2asin0)2 + (acosfl)2

(v)

(vi)

(vii)

(viii)

(ix)

(x)

Sec. 6.6 Torque

Fv = Fsind) = FL = - z

* r y r V(2asin0)2 + (a cos 0)2

213

(xi)

2-a-sin(9)-2-a-cos(0)

a-J(2-a-sin(9))

-k-(-a-sin(9))-de

(2-a) -H(a-cos(9))

Substituting these inEq. (viii), the work Wb W b _ _k.ione in going from A toC by the circular path is j 0

>̂ shown.

Which is the same as in Part(a) Wb=-k-U5- l + ln(2)-

EXERCISE 6.3 Repeat the problem if the force directed toward the origin is F = klx.

i 6 TORQUE

Let us consider a particle of mass m located at point P at a distance r from the origin and acted•n by a force F, as shown in Fig. 6.15, both r and F being in the XYplane. We want to calculate

T- about an axis passing through 0 and perpendicular to the XY plane. The torque or moment of- <rce about the origin O is defined as the product of the distance r (= OP) and the component'f force F perpendicular to r; that is, F sin cf) = F sin(7r — ff) — F sin 6. Hence

rn = rF sin 0 or Tn = r x F (6.80)

The torque T0 will be along the +Z-axis if F acts counterclockwise and will be along the —Z-axis:f F acts clockwise.

Let us generalize the preceding definition of the torque as applied to a three-dimensionalcase. As shown in Fig. 6.16, the force F acts on a particle at P that is at a distance r from theorigin O. We want to calculate the torque or the moment of force of F acting at P about an axis

Figure 6.15 Torque T0 due to a forceX F at a distance r from the origin O.


Figure 6.16 Torque TNN> (in a three-dimensional case) due to a force F at adistance r from the origin O.

AW passing through O. Let us resolve F into two vector components: Fy, a component parallelto AW' and F±, a component perpendicular to AW. Thus

F = F, + F, (6.81)

Since n is a unit vector along the axis AW, n • F is the projection of F along AW'; hence theparallel component is

while the perpendicular component is

FM = n(n • F)

= F - FM

Thus the torque about the axis AW may be defined as

TNN, = ± rF± sin 0 = ± |r X F j

(6.82)

(6.83)

(6.84)

The positive sign is for the torque along n and the negative sign for the torque opposite to n.On the other hand, the component F|| to AW' does not produce any torque along the AW axis.Since r x Fn is a vector that is perpendicular to n, its component parallel to ii will be zero;that is,

n • (r x F||) = n|r x Fyl cos 90° = 0

This result enables us to write TNN, given by Eq. (6.84), as

TNN, = ± | r x F j = n -(r x F)

We can prove this result as follows:

n • (r x F) = n • (r x (F,, + F J ]

(6.85)

(6.86)

Sec. 6.7 Dynamics in Three Dimensions 215

= n • (r x F||) + n • (r x F±)

= n • (r x F±)

Thus, without considering the orientation of the position vector r and the applied force F, the:orque about the axis AW with r being drawn from any point from the axis is given by

= n • (r x F) (6.87)

This automatically gives the correct sign and takes care of the Fy component, which makes nocontribution. We may consider TNN> given by Eq. (6.86) to be a component of T0, which is de-nned by

T0 = r X F (6.88)

•vhere T0 is the torque about an axis passing through 0 and r is the distance from O to P.

i 7 DYNAMICS IN THREE DIMENSIONS

The general equation describing motion in three dimensions may be written as

d2r

atv> 0 (6.89)

•A hich is a set of three coupled simultaneous second-order differential equations and may be.vritten explicitly as

d2xm —y = Fx(x, y, z, x, y, z; t) (6.90)

•vith similar expressions for y and z components. Thus, if the initial position ro(xo, y0, z0) andJie initial velocity vo(%, v0 , vOz) are given, we know the six arbitrary constants x0, y0,L . vOx, v0 vOz in the form of initial conditions and we can proceed to solve Eqs. (6.90). Solving•Jiese equations is a difficult task and is usually carried out by using numerical analysis.

There are situations where it is possible to solve these equations for two- or three-dimensional motion. For example, if each force component given depends only on the corre->ponding coordinate and its derivative, Eqs. (6.90) take the form

Sx(6.91a)

m

m

dt2

di2

= Fy(y, y, t)

= FJz, z, t)

(6.91b)

(6.91c)

These are three independent one-dimensional problems and can be solved for x(t), y(t), and z{t)by familiar methods used in previous chapters. A simple example is that of three-dimensional

216

1Motion in Two and Three Dimensions Chap. 6

y

Figure 6.17 Harmonic oscillation inthree dimensions with spring constantkx, ky, and kz.

harmonic oscillator, such as the motion of an atom in a crystal lattice with a cubic structure,shown in Fig. 6.17. The forces in this case are

Fy = - (6.92)

which involve solving separately for three linear harmonic oscillators.A few remarks about the potential function and the total energy are in order. In the case

of one-dimensional motion, if force is a function of position only, that is, F = F(x), we can al-ways define a potential energy function

V(x) == - f F(x)X.

dx (6.93)

This is true because, when the particle moves from Xj to x2, it has no choice but to return by thesame path; hence the work done in a round trip is zero. Thus the total energy is constant, K +V = E, and the energy integral can be used to solve the one-dimensional problem.

In the case of three-dimensional motion, even if the force is a function of position only,that is, F = F(r), it does not guarantee the existence of a potential energy function V(r). Whensuch a potential function does exist, the conservation of energy theorem still holds, that is, K +V = E = total energy. But unlike the case of one-dimensional motion, the energy integral is nolonger sufficient to solve a problem of two- or three-dimensional motion.

Sec. 6.8 Harmonic Oscillators in Two and Three Dimensions

5 8 HARMONIC OSCILLATORS IN TWO AND THREE DIMENSIONS

217

We extend our discussion of the previous section to the motion of two- and three-dimensionalharmonic oscillators in this section and projectile motion in the next section.

Harmonic Oscillations in Two Dimensions:Lissajous Figures

A typical example of a two-dimensional anisotropic oscillator is shown in Fig. 6.18. We shalllimit our discussion to anisotropic oscillator for which kx = ky = k. If the restoring force is pro-portional to the distance,

F = - A T

using polar coordinates, we can resolve F into two components as

Fx = —krcos 6 = —kx

Fy = — kr sin 6 = — ky

Hence the differential equation, Eq. (6.94),

may be written as

m —T- = —krdt2

x + w2x = 0

y + (o2y = 0

(6.94)

(6.95)

(6.96)

(6.97)

(6.98)

(6.99)

'••hhhhhhh tt N

s-

y -£.

Figure 6.18 Two-dimensionalanisotropic harmonic oscillator.

where co2 = him. The solutions of these equations are

x = A cos(&>? + 4>x)

y = B cos(citf + <f> )

(6.100)

(6.101)

where A, B, (f>x, and <py are constants. Both oscillators have the same frequency, but may havedifferent amplitudes and phases. We can obtain the path of the particle by eliminating t fromEqs. (6.100) and (6.101). We may write Eq. (6.101) as

- = cos[a>f {<t>y ~

= cos(wt + <f>x) cos((f)y — (t>x) — sin(w? + cf}x) sin(^>>. — (f>x)

Substituting for cos(wf + <f>x) and sin(wf + <frx) from Eq. (6.100) and rearranging,

- - - cos(<f>y - ^ )

Squaring both sides and rearranging, we get

X

X

A22xy

(6.102)

This is an equation of an ellipse, as shown in Fig. 6.19. The major axis of the ellipse makes anangle \\i with the X-axis and is given by (see Problem 6.30)

tan 2ijj =2AB

(6.103)A2-B2

[Note: Eq. (6.102) is a general equation of the form

ax2 + bxy + cy2 + dx + ey = f

which represents an ellipse if the descriminant b2 — Aac is negative, a hyperbola if the discrim-inant is positive, and a parabola if it is zero.]

A' /

<

Y )

\

\

B

\

>

/A X

B'

Figure 6.19 The resultant motion oftwo simple harmonic motions at rightangles to each other, in general, is anellipse.

Sec. 6.8 Harmonic Oscillators in Two and Three Dimensions 219

If, in Eq. (6.102), <f>y - 4>x = TT/2, we get

? ? (6.104)

which is an equation of an ellipse with its major and minor axes coinciding with the X- and Y-axes, as shown in Fig. 6.20. On the other hand, if <f> — <j>x = 0 or 77, Eq. (6.102) reduces to

By = ± — x

A

These equations represent straight lines.Considering Eq. (6.104) again, if A = B, it takes the form

x2+y2=A2

(6.105)

(6.106)

which is an equation of a circle. As a matter of fact, the shape of the orbit changes as the phasedifference cf) = <f>y — 4>x changes its values, as illustrated in Fig. 6.21 for the case when A = Band 4> = 0°, 30°, 6 0 ° , . . . .

Let us now consider the case in which the two frequencies are not equal; hence the solu-tions of Eqs. (6.98) and (6.99) are of the form

x = A cos(wxt + <f>x)

y = B cos(a>yt + (f> )

(6.107)

(6.108)

The path of the particle is no longer an ellipse. The paths obtained in such cases are Lissajouscurves or figures. Depending on the ratio of u>Ja>y, the curves may be open or closed, as ex-plained next. Figure 6.22 illustrates the Lissajous figures for the case coy = 2cox and <j> = 0, TT/4,77/2, . . . . There are many experimental arrangements, both mechanical and electrical, thatdemonstrate Lissajous figures.

Harmonic Oscillators in Three Dimensions

As mentioned earlier, the forces acting are such that each component of force is a function ofthe corresponding single coordinate. A typical example is the motion of an atom in a crystal lat-tice with a cubic structure. We assume that there is no damping and that the spring constants arekx, k , and kz. The situation is as shown in Fig. 6.17. Thus the equations describing the motionsof three one-dimensional harmonic oscillators are

Fx = mx = —k^

Fy = my = -kyy

Fx = m'z = -k/:

The corresponding equations resulting in three independent motions are

x = Ax cos(to/ + (f>x)

(6.109a)

(6.109b)

(6.109c)

(6.110a)

Figure 6.20

As we know, the resultant motion of two simple harmonic motions at right angles to eachother, in general, is an ellipse. If the two amplitudes are equal, the ellipse becomes a circle.

Fx and Fy are componentsof force F.

The equations describingthe two simple harmonicmotions are

The equations describingdisplacements, velocities,and accelerations for thetwo oscillators are

F=-kr Fx=-k-r-cos(9)=-k-x Fy»-k-r-sin(9)«-k-y

x=0d 2 2

y-t-co y=0

x=A-cos(co-t-i-<|))

vx=-A-sin( co-1-H <|)) • <n

ax=- A- cos( co-1 -\- <|)x) • co

2 kCO = —

m

N:=50 t : = 0 . . N A:=30

y=B-sin(co-t-i-<j>)

vy=B-cos(co-t-i-(|))-co

2

ay=-B-sin( co-1 •+- «j>y) • co

B :=60 co:=.4

c|)xl :=0 (|>yl :

l ( :=A-cos(co-t-i-())xl)

l ( := Asin(co-

0x2 :=0 <j>y2 :=%

x2( :=A-cos(co-

y2 :=B-sin(co-t-)-(|)y2)

(a) In both graphs the major andminor axes coincide with the xand y - axes. What will cause ashift in the position of the major orminor axis?

(b) In both graphs <j)yl - 0x1 = 0or 7i. If different values are used,how will the graphs differ?

30

0

t

30

—AnV

J-60 "30 30 60

xlt,x2(

Figure 6.21

More accurately, the resultant motion of two simple harmonic motions at right angles toeach other can result in any one of many graphs, depending on the relative amplitudesand the phase differences between the two motions: a circle, an ellipse, an orientedellipse, or a straight line.

co = frequency of oscillation, _which is the same for both simpleharmonic motions.

A and B are the amplitudes ofthe two simple harmonic motions.

t : = 0. .N i :=0..6 co :=.4

A:=30 B : = 60 <bx. :=0 <b;

x . :=A-cosfco-t-t- <bx.) y . :=B-cos(co-t-i-<])y.

71

6

Ox. and (by. are the phase angles, x and y are the displacements of the two simple harmonic

motions.

100

50

y t , 0 0

- 40 -20 0 20 40

X

t 0

YX0 = 0 <by0 = 0 -deg

100

50

yt,i

"50

-100-40 -20 0 20 40

Xt,l

j)X]=0

100

50

0

-50

-100

= 0 (by =60'deg

-40 -20 0 20 40

Xt,2

100

50

y t ,3 0

-50

-100- 40 -20 0

\3

20 40

Yx = 0 Ty =90»deg

Figure 6.21 (continued)

100

50

yt,4 0

-50

-100

0X

50

yt,6 0

\\

^

— •

\

\

50

y t , 5 0

-40 -20 0 20 40 1°°-40 -20 0 20 40

4 = 0 <|)y4 = 120*deg

^ \

\

-40 -20 0 20 40

X

t 6

(|)x = 0 <j>y = 1 5 0 #deg

The following figure shows all theabove graphs together.

Which of the above graphs willbecome circle, if A = B?

i <))x

0

I2

3

4

16

0

0

0

0

00—0

180

n0

0.5241.0471.5712.094

2.618

3.142

0

306090

120

150

180

-30 "20 "10 ;0 30

Figure 6.22

I :=100 i :=0 . . I j :=0..4 t. := —A Lissajous figure results from the 10

combination of two simple harmonic A : = 1 B : = 1 cox := l coy:=2

motions at right angles, with . /° x. . : = A-cos(cox-t.-t-<

different frequencies and different 1J V 'amplitudes, as shown here. „

y. j : = B-c3-*

1 17

Ai,0

yi,2

0

L i , 2

f)1 /

\ \\ \

0

Xi,3

-10

x. .1,4

y = Aycos(a)yt + <f>y)

z = Az COS(OJJ + 4>z)

where cox, w , and coz are the angular frequencies given by

2 x

m}'

mKm

(6.110b)

(6.110c)

(6.111)

The six constants Ax, Ay, Az, <f>x, <j)y, and <f>z depend on initial conditions: the initial coordinatesx0, y0, andz0 and the initial velocities vQx, vQv, and vOz. The resulting motion of the particle isconfined to a rectangular box of dimensions 2AX X 24 X 24Z placed about the origin.

If the angular frequencies are such that for some set of integers nx, n , and nz

^ = Mj = ^ (6.112)nx ny nz

the frequencies are said to be commensurable. The motion of mass m in such cases is closed,and the motion repeats itself at regular intervals of time. Furthermore, if the set of integers nx,nY, and nz is such that they have no common integer factor, then the period of motion is given by

_coy

(6.113)

This means that, in one time period T, coordinate x makes nx oscillations, coordinate y makes ny

oscillations, and coordinate z makes nz oscillations. At the end of one period, the particle returnsto its initial position and velocity.

If (ox, coy, and coz in Eq. (6.112) are incommensurable, the curve describing the motion ofthe particle will never pass through the same point twice with the same velocity. The motion isnot periodic. The path fills the entire box of volume 2AX X 2A X 2AZ, and after a sufficientlylong time the particle will eventually pass arbitrarily close to every point in the box.

Let us discuss an interesting example of a three-dimensional isotropic harmonic oscilla-tor; that is, all the spring constants are the same and so are the frequencies:

= k,, = k7 and a> = —y z

m

(6.114)

Therefore, x — Ax cos(cot + 4>x)

y = Aycos(a)t + 4>y)

z = Az cos(arf + <t>z)

(6.115a)

(6.115b)

(6.115c)

To determine the path or the trajectory of the particle we eliminate t and express z in terms of xand y. The result is

z = K,x + K7y (6.116)

Sec. 6.9 Projectile Motion 225

(see Problem 6.40) where K{ and K2 are constants. This is the equation of a plane. Irrespectiveof initial conditions, the motion of an isotropic harmonic oscillator is always confined to a plane.The motion is periodic, and each coordinate executes one cycle of oscillation in each period.The path of the particle can be shown to be either an ellipse, a circle, or a straight line.

6 9 PROJECTILE MOTION

To begin, we limit our discussion to two dimensions: without and with air resistance.

No Air Resistance

Consider a projectile of mass m that is launched from the origin of a coordinate system with ve-locity v0, making an angle a with the horizontal axis, as shown in Fig. 6.23(a). The only forceacting is the downward force of gravity, and the motion of the projectile in the XZ plane is de-scribed by the equation

m df = -mgt.

or. in component form,

d2x d2zm —r = 0 and m —r = — mgdt2 dt2

(6.117)

(6.118)

Assume that the starting point is the origin (0, 0), and the initial horizontal and vertical compo-nents of velocity v0 are

X0 = ~ V0 C O S a z 0 = Vz0 = v0 S m a

Using these initial conditions, the solutions of Eqs. (6.118) are, after integration,

x = vx = vM = constant

z = vr = v'zO

(6.119)

(6.120)

(6.121)

Figure 6.23(a) A projectile launchedfrom the origin of a coordinate systemwith velocity v0 and making an angle awith the horizontal is shown. zm is themaximum height and R is the range.

226

and, after a second integration,


X =

z =

(6.122)

(6.123)

The path of the projectile is obtained by eliminating t between Eqs. (6.122) and (6.123), result-ing in the trajectory

z = ^ x -2trn

(6.124)

which is an equation of a parabola, as shown in Fig. 6.23(b).Next we proceed to find the range R. Setting z = 0 in Eq. (6.124), we obtain two values

of x: x = 0, corresponding to the starting position, and x = R, given by

R =; cos a sin a sin 2a

g g 8This equation indicates that, for a given value of v0, R will be maximum when a = 45°.

\ Figure 6.23(b)

(6.125)

The path of a projectile that is launchedwith a velocity of 33 m/sec and makingan angle of 25 degrees with thehorizontal is shown.

I :=30 i :=0..I t. : = —-sec1 10

g:=9.8-— 6:=25deg

vxO: = vO-cos(9)

vzO :=vO-sin(0)

x. :=vx0t.

= 29.908'm'sec- l

Note that the projectile travels vertically9.92 m before landing at a horizontaldistance of 89.72 m.

x30 = 89.724'm

max(x) = 89.724'm

max(z) =9.92 I'm

vzO = 13.946'tn'sec

vzOz. := -x. -

1 vxO '

- l

2-vxO

10

7.5

5

2.5

00 25 50 75 100


The maximum height zm reached by the projectile, as shown in Fig. 6.23(a), is obtained byrewriting Eq. (6.124) as (see Problem 6.42)

x — z —vz0 (6.126)

Since z = zm, when [from Eq. (6.125)] x = R/2 = v^vjg, Eq. (6.126) gives

Air Resistance as a Function of Velocity

(6.127)

Let us assume that air resistance varies linearly with velocity. Since air resistance always op-poses motion, the direction of the resistive force is in the direction opposite to that of v. Thusthe equation describing the motion is

cfL_

dr

where b is a constant of proportionality for the resistive force and

r = n + zk

v = xi + zk

(6.128)

(6.129)

(6.130)

(6.131)

where x = uxandz = vz. Thus Eq. (6.128) may be resolved into the following two components:

mx= -bx (6.132)

x'z = -mg - b'z (6.133)

These equations can now be integrated by methods familiar to us from solving one-dimensionalproblems. Thus, assuming that at t = 0, (x0, z^) = (0, 0), and u0 = (x0, z0) and integratingEqs. (6.132) and (6.133), we obtain

x =xoe-h"m

and, integrating again,

2^

fe2 fe r

(6.134)

(6.135)

(6.136)

(6.137)


Combining Eqs. (6.134) and (6.135), the expression for the velocity of the projectile is

v = xi + zk

(mg= xne

"IKb

+ zn\e^-btlm

For large values of t, this velocity approaches the terminal velocity:

mgvt = , for large values of t

And from Eq. (6.136) for large values of t, we obtain the limiting value of x to be

for large values of tX'= b>

(6.138)

(6.139)

(6.140)

Also from Eq. (6.137) for large values of t, z approaches minus infinity. All this leads to the fol-lowing conclusion: The trajectory has a vertical asymptote, the line x = xom/b; that is, it has avertical drop, as shown in Fig. 6.24(a) and (b) and this drop begins above the horizontal plane

z = 0.Let us now compare the results obtained in Eqs. (6.136) and (6.137) with those in the case

of no air resistance. By using the exponential series

" + 2T+3Twe may write Eqs. (6.136) and (6.137) as

bx0

2m

1

zk

b / z ¥ gt3

m+ •

(6.141)

(6.142)

Verticalasymptote

Figure 6.24(a) The trajectory of aprojectile with and without air resis-tance, where the air resistance is pro-portional to the velocity.


Figure 6.24(b)

\0 and zO when no air resistancex and z when there is air resistance

N:=30 i:=0..Nb = 0.2 is the constant of

proportionality for the resistive force. b ,_ 2 m •=5 vxO =100The initial velocity in both the x and zdirection islOO m/sec. b t i

The maximum values of x and z arecalculated below.

(a) How does the value of b affect \ '•=- — (ti-the maximum values of x and z?

m'-g m-vzO

(»>) Graph for the values of b to be0.1 and 0.4 and discuss (a).

°^ vzOti ( J ' g '

t. :=i

vzO :=100

xO. :=vx0-t.

g:=9.8

-b-t.

1-e

(c) If the resistance is proportional tothe square of the velocity, how willthis affect the maximum values?

max(x) = 1.747-103

max(xO) =3-103

max(z) =402.542

max(zO) =510

max(zO) - max(z) = 107.458

max(xO)- max(x) = 1.253-103

1000

z0.

-1000'3000

Projectile with, without air resistance

The first term on the right in Eq. (6.141) and the first two terms on the right in Eq. (6.142) arethe same as in the case of no air resistance. The remaining terms in each case are the correctionterms, which are very small for the case in which btlm is small as compared to unity.

Finally, let us obtain an equation of the trajectory by eliminating t between Eqs. (6.136)and (6.137). This yields

m2gIn mxn

b2 \mx0 — bx(6.143)

For low air resistance or over short distances, that is, for (bx/mx0) <t 1, we may write the lastterm as ln(l + u), expand it, and obtain

Zn 1 g -, 1 bg ,z — . x .2 x

XQ 2Xn 3(6.144)

This equation without the last term in the equation of a parabola, while the last term is a cor-rection term. That is, Eq. (6.144) represents a parabolic trajectory with a small correction term.As before, we can obtain the maximum range by substituting x = xm when z = 0. We obtain

=

2 b

3 i m (6.145)

The first term on the right is the range R when there is no air resistance, while the second termis the correction term. To solve Eq. (6.145) for xm we can use the approximate valuexm — ix^Jg in the last term and thus obtain

(6.146)mg

On the other hand, if air resistance is large and is a major factor in determining the range—thatis, if (bZfJmg) > 1—the maximum range is (see Problem 6.46)

mxn (6.147)

Projectile Motion in Three Dimensions

For no air resistance and air resistance linearly proportional to the velocity, we do not get anydifferent results from that obtained in the case of two-dimensional projectile motion. An inter-esting situation is one in which there are crosswinds resulting in a drift force Fd in the Y direc-tion, as shown in Fig. 6.25. The resulting equations of motion are

d2rm

dr-^ = - mgk - b\ + FJ

(6.148)

z,.

Figure 6.25 Drift force Frf due to acrosswind in the Y direction acting on aprojectile with an initial velocity v0 inthe XZ plane.

Problems 231

which in component form may be written as

mx = — bx

my = — by + Fd

rriz = —bz — mg

(6.149)

(6.150)

(6.151)

These may be solved as before, yielding almost the same results. We leave this as an exercise(Problem 6.58).

PROBLEMS

6.1. Derive Eq. (6.41), an expression for acceleration in cylindrical coordinates.

6.2. Derive the relations given in Eq. (6.47) from geometrical considerations.

6.3. Starting with Eq. (6.49) for velocity in spherical coordinates, derive an expression for accelerationgiven by Eq. (6.50).

6.4. The motion of a particle is described by the following equations. Find the velocity and accelera-tion, giving a geometrical interpretation, if any, in each case.(a) r = a& + bt2]

(b) r = ati + A cos cot j(c) r = ati + A cos wfj + B sin &>?k(d) r = e*'ur, 0 = at, in polar coordinates(e) r = aur, 6 = B sin cot, 4> = bt, in spherical coordinates.

6.5. Consider a river of width w. The speed of water near the banks is zero, but increases linearlyand reaches a value vc at the center of the river. If a boat starts straight across from one bankwith a speed vb, show that when it reaches the other bank, it has drifted downstream by a dis-tance vcw/2vb.

6.6. Show that the path of a particle is an ellipse having a major axis of 2a if its position vector is givenby r = (a sin cot)i + (b cos a>t)\, where a, b, and w are constants. Also calculate the speed of theparticle.

6.7. A particle moves with a constant speed v in a parabolic path given by y2 = 4fx, where/is a con-stant. Find its velocity and acceleration components in rectangular and plane polar coordinates.Show that the equation of the given parabola in plane polar coordinates is

1 - cos 6 or 20 fr cos = f2

6.8. The acceleration a of a particle is a function of time t. Find the r and 8 components of da/dt in planepolar coordinates.

6.9. A vector A represents the position of a moving particle and is a function of time. Find the compo-nents of d2A/dt2 in cylindrical polar coordinates.

6.10. A vector A represents the position of a moving particle and is a function of time. Find the compo-nents of d3A/dti in spherical coordinates. (The time derivative of acceleration is called the jerk.)

6.11. The rate of change of acceleration is defined as the jerk. Find the magnitude and direction for thejerk of a particle moving in a circle of radius R and angular velocity co.

6.12. A particle is moving with constant speed, but with continuously changing direction. Show that theacceleration vector is always perpendicular to the velocity vector. In general, if a vector A has con-stant length A but its direction in space is changing with time, then the rate of change of A is an-other vector that is always perpendicular to vector A.

6.13. Calculate the forces corresponding to the following potential functions: (a) V = ax2 + by2 + cz2,and (b) V = Kye ~x, where a, b, c, and K are constants.

6.14. Are the following forces conservative? Find the potential energy function if it exists.(a) F = Kxyzft + j + £), where K is a constant.(b) F = K(n + yj + zk), where A" is a constant.(c) F = K(x2yi + xy2} + x2y2&)caz, where K and a are constants.

6.15. Show that the following force is conservative and find the corresponding potential.

Fr = axe" = cze

where u = ax2 + by2 + cz2 and a, b, and c are constants.6.16. Which of the following forces are conservative? If any, find the corresponding potential.

(a) F = ax2! + ay2] + az2i.(b) Fx = 3az(x2 - y \ Fy = -6azxy, Fz = ax(x2 - 3y2)(c) Fx = -ay2, Fy = ayz, F7 = -ay2

6.17. Determine which of the following forces are conservative. Find the potential energy function forthose that are conservative.(a) F = ax2yzi + bxy2z\ + cxyz2k(b) F = cx2yA + cxy2zj + cxyz2k(c) F = ax2vl + ay2xj + az3k

6.18. Show that the following force is conservative and find the correspondiag potential.

Fr = ar2 cos d, Fe = ar2 sin 0, Fz = laz2

where a is a constant.6.19. Show that the following force is conservative and find the corresponding potential.

Fr= — 2ar sin 9 cos <$>, Fe= — ar cos 6 cos 4>, V^ = ar sin-Sfiln </>where a is a constant.

6.20. Consider a particle of mass m in a uniform gravitational field F = -mgk. Calculate the work donein moving from point A to C along the three paths shown in Fig. P6.20.(a) From A to B and then from B to C—path (J).(b) From A to O and then From O to C—path (2).(c) Along the quadrant of a circle from A to C—path ®.

Y ,

•jro ai

©'

0

r,

© B

\1

© r/ =

©

c= o! x

Figure P6.20

Problems 233

6.21. Repeat Problem 6.20 if the particle moves in a force field given by F = axyi + bx2yj.

6.22. Calculate the work done by a particle of mass m as it moves along a straight line from the origin(0, 0, 0) to the point (x0, y0 z0) under the following force:

Fx = ax1 + bxy + cxz, Fy = ay2 + bxy + cyz, Fz = az

6.23. A body of mass m can be moved from point A to point C along three different paths shown inFig. P6.23:

Yi

o(OtO)

C(4, 4)

D(4, 1)

Figure P6.23

(a) From A to B and then from B to C.(b) From A to.D and then from D to C.(c) FremA to C directly.

Calculate the work done in each of the three cases for the following forces:(i) F = ax\. + by}

(ii)F = (x+:a%+(y + b)j(iii) F = rH>i, + yhc](iv>F —xA \

6.24. Calculate the work done in moving a body of mass m from point A to point C along three differ-ent paths, as shown in Fig. P6.24:

A(0, a) B(a, a)

C(a, 0)

(a) From A to B and then from B to C.

Figure P6.24


(b) Along the straight line path directly from A to C.(c) Along an arc of a circle from A to C.

Carry out these calculations from each of the following three forces:(i) F = fcd + ky]

(n)F = - 1 + - Jx y

(iii) F = xTi + yTi6.25. Calculate the work done in moving a body of mass m from point A to point C along two different

paths, as shown in Fig. P6.25:

Y ,

A(0, a)

O(0,0)

~ ^

B(2a, a)

C(2a, 0) XFigure P6.25

(a) From A to B and then from B to C.(b) Along a segment of an ellipse from A to C. Use the relations x = 2a sin 6 and y — a cos 6.

Carry out these calculations for each of the following three forces:(i) F = fcd + kyj

(ii) F = 2yi - 2x\

(iii) F = x2yi + xy2j

6.26. A particle of mass m moves such that

x = x0 + at2, y = z = ct

Find angular momentum L as a function of time t. Also find F and T and verify that the angularmomentum conservation theorem is satisfied.

6.27. At time t = 0, a force F = 2ai + 3b) is applied to a particle of mass m at rest at the origin. Findits position and velocity as a function of time.

6.28. At time t = 0, a force F = a& + bi1) + cf3£ is applied to a particle of mass m at rest at r0 = 2i +3 j . Find its position and velocity as a function of time t.

6.29. At time t = 0, a force F = aR +bek'J is applied to a particle at r0 = 2i +3j and moving with ve-locity v0 = vMi + Vyoj. Find the position and velocity of this particle as a function of time t.

6.30. Derive Eq. (6.103).

6.31. Consider two simple harmonic motions:

x = A cos coxt, y = A cos(u>yt + 4>)

Draw Lissajous figures if u>y = 2u>x and 4> = 0, TT/4, TT/2, and TT.

Problems 235

6.32. Repeat Problem 6.31 if 2coy = 3cox.

6.33. Two harmonic vibrations at right angles to each other are described by the equations

x = 12 cos(4Trt), y = 12COSI8TTT + -

Draw Lissajous figures that describe this motion.

6.34. Draw Lissajous figures that describe the following motion:(a) x = 4 sin 2cot and y = cos 2iot(b) x = 5 cos cot and y = 5 cos 2cot

6.35. Draw Lissajous figures for the following motion:(a) x = 5 sin wt and y = 5 cos 2u)t(b) x = 5 cos(2wr) and y = 5 cos(2w? + TT/4)

6.36. Draw Lissajous figures for the case <f>x = (py and Aa>x = 3coy.

6.37. Draw Lissajous figures for the case <f>x = <f)y and 3cox = 4coy.

6.38. Consider a nonisotropic two-dimensional harmonic oscillator for which the potential energy is

V(x, y) = [x2 + \y2

Solve for the equations of motion, considering the particle to have a unit mass. Let r0 = a j a n dv0 = a , i . Draw an orbit representing this motion.

6.39. Consider a two-dimensional isotropic harmonic oscillator of mass m represented by

x(t) = A cos cot and y(t) = B sin cot

Show that under the appropriate initial conditions and proper coordinate system, the angular mo-mentum L and total energy E are

L = VkmAB and E = \k(A2 + B2)

What is the expression for r in terms of L, E, £,and of!

6.40. Derive Eq. (6.116) for a three-dimensional isotropic oscillator.

6.41. Discuss the motion of a three-dimensional oscillator for which the motion is confined to a box ofdimensions 2A X 2B X 2C.

6.42. Write Eq. (6.124) in the form given by Eq. (6.126).

6.43. Starting with Eqs. (6.132) and (6.133), derive Eqs. (6.134) to (6.137).

6.44. Show that, for large values of t, Eq. (6.138) reduces to Eq. (6.139) and Eq. (6.136) reduces toEq. (6.140).

6.45. Starting with Eqs. (6.136) and (6.137), derive Eq. (6.143).

6.46. Starting with Eq. (6.143), derive Eqs. (6.146) and (6.147).

6.47. Derive an expression for xm similar to the one given in Eq. (6.146), but keep an extra term in theexpansion arrived at in Eq. (6.145).

6.48. Consider a projectile fired from the origin with velocity vo(iiOx, vOy, vOz) in the presence of a windvelocity vw = vwj and air resistance proportional to the velocity. Solve the equations of motion forthe coordinates (x, y, z) as a function of time t. Suppose the projectile returns to the horizontal planeat a point (xh y{). Calculate the position keeping only the first-order term in b. Show that, if we ne-glect air resistance, the target distance is missed by a fraction 4bvz0/3mg of its target distance, andthe wind causes an additional miss in the y coordinate by an amount 2bvuv

l2SJ{mg2).

6.49. Using the equations in Problem 6.48, make the appropriate plots using numerical values.

6.50. A projectile is fired from the origin with velocity vQ, making an angle 6 with the X-axis in the XZplane, and hits the target at (xQ, 0). Find the first-order correction to the angle of elevation due toair resistance.

6.51. Using numerical values, make the graphs in Problem 6.50.

6.52. A gun fires a shot with a velocity v0. Calculate the maximum range in any direction, given that theangle of elevation a of the gun may be varied. From Fig. P6.52 show that

r = Y~- cos a sin(a — 9)e

and<"-« g ( 1 + s i n e)

The maximum range Re is in a direction making an angle 6 with the horizontal.

z A

Figure P6.52

6.53. Using numerical values, make the graphs in Problem 6.52.

6.54. A gun mounted at the foot of a hill when fired strikes the hill at a right angle. If the hill makes anangle (j> with the horizontal, find the angle that the gun barrel makes with the slope of the hill.

6.55. Two projectiles are fired, one with velocity v0l making an angle 6{ and the other with velocity v02

making an angle d2 (6, > 62). Show that if they are to collide in midair the time interval betweenthe two firings must be

2vmvQ2sm(6l - 02)

g(vm cos v02 cos

6.56. A projectile is fired with velocity u0 and passes through two points, both a distance h above thehorizontal. Show that, if the angle of the barrel of the gun is adjusted for the maximum range, thenthe horizontal separation of the two points is

6.57. The motion of a falling spherical drop of liquid of density p0 is opposed by (a) a force proportionalto the surface area of the drop, (b) the density p of the medium, and (c) the nth power of its veloc-ity. Show that the terminal velocity varies as the wth root of the radius of the drop. Using appro-priate values, graph v versus t and v versus x.

6.58. Solve Eqs. (6.149), (6.150), and (6.151).

Suggestions for Further Reading 237

SUGGESTIONS FOR FURTHER READINGARTHUR, W., and FENSTER, S. K., Mechanics, Chapters 2, 5, and 7. New York: Holt, Rinehart and Winston,

Inc., 1969.

DAVIS, A. DOUGLAS, Classical Mechanics, Chapters 5 and 6. New York: Academic Press, Inc., 1986.

FOWLES, G. R., Analytical Mechanics, Chapters 2 and 4. New York: Holt, Rinehart and Winston, Inc., 1962.

FRENCH, A. P., Vibrations and Waves, Chapter 2. New York: W. W. Norton and Co., Inc., 1971.

HAUSER, W, Introduction to the Principles of Mechanics, Chapter 2. Reading, Mass.: Addison-WesleyPublishing Co., 1965.

KITTEL, C , KNIGHT, W. D., and RUDERMAN, M. A., Mechanics, Berkeley Physics Course, Volume 1, Chap-ter 5. New York: McGraw-Hill Book Co., 1965.

ROSSBERG, K., Analytical Mechanics, Chapters 1 and 2. New York: John Wiley & Sons, Inc., 1983.

SVMON, K. R., Mechanics, 3rd ed., Chapter 3. Reading, Mass.: Addison-Wesley Publishing Co., 1971.

c h a t r motion in two and three dimensions · in describing motion in two and three dimensions,...

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